Chapter 5 Chemical Equilibrium

Chapter 5Chemical Equilibrium

2

The Equilibrium State

Learning objectives: Define chemical equilibrium State examples of chemical equilibrium Write a balanced chemical equation for any reversible

reaction. Interpret how equilibrium state is achieved from a

reaction.

3

The Equilibrium State• What does equilibrium mean to you?

• Chemical Equilibrium: A state achieved when the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant.

– Examples are . . .

• Saturated solutions

• Phase equilibrium

• Many reactions

• Human body equilibria involving O2 molecules and the protein Hb play a

crucial role in the transport and delivery of O2 from our lungs to cells throughout our body.

4

Equilibrium between phases is known as physical equilibrium.

The Equilibrium State• Chemists are interested in these reversible reactions.

One example is the following:

The decomposition of the colorless gas dinitrogen tetroxide (N2O4) to the dark brown gas nitrogen dioxide (NO2).

5

N2O4 (s)Frozen N2O4is nearly colorless

N2O4 (g) 2NO2 (g)Eventually the color stops changingas N2O4 (g) and NO2 9g) reachconcentration at which they areinterconverting at the same rate. The wo gases are in equilibrium.

N2O4 (g) 2NO2 (g)As N2O4 is warmed above its boiling point,it starts to dissociate into brown NO2 gas


6

2NO2(g)N2O4(g)The Equilibrium State

Change in the concentrations of N2O4 and NO2 with time in two experiments at 25°C

8

Important lesson about equilibrium…

A mixture if reactants and products is formed in which concentrations no longer change with time indicates that the reaction has reached a state of equilibrium.

For equilibrium to occur, neither reactants nor products can escape from the system.

At equilibrium, the particular ratio of concentration terms equals to constant.


9

The Equilibrium Constant

Learning objectives: Write equilibrium constant expression. Calculate the equilibrium constant Kc from the

equilibrium concentrations of products and reactants.

10

• Equilibrium expression for determining the equilibrium constant, Kc, for the general reversible reaction

aA + bB cC +dD.

The Equilibrium Constant Kc

In this expression: Product concentrations always appear in the numerator Reactant concentrations always appear in the denominator Each concentration is always raised to the power of its stoichiometric

coefficient in the balance equation Kc is independent of concentration changes, but dependent on the

temperature.

The Equilibrium Constant Kc

= 4.63 x 10-3

0.0429

(0.0141)2

Experiment 5

[N2O4]

[NO2]2Kc = = 4.64 x 10-3

0.0337

(0.0125)2

Experiment 1

Kc for a chemical reaction is very useful because it indicates whether a reaction is product or reactant favored, and it can be used to calculate the quantity of reactant or product present at equilibrium

12

Homogeneous &Heterogeneous Equilibria

Learning objectives: Distinguish between homogenous and heterogeneous

equilibrium.

Write the equilibrium equation for any balancedchemical equation representing a homogenous orheterogeneous equilibrium.

Calculate the equilibrium constant Kp from the equilibrium partial pressures of reactants and products.

Interconvert between Kc and Kp using a balancedequation.

13

• Homogeneous Equilibrium: When all reacting species are in the same phase, all reactants and products are included in the expression.

• Eg.

Homogeneous Equilibria

Ag+ (aq) + 2NH3 (aq) Ag(NH3)2+ (aq)

diamminesilver(I) cation

Co (g) + Cl2 (g) COCl2 (g)phosgene

14

• Kc is obtained when equilibrium concentrations expressed in molarity are substituted into equilibrium –constant expression.

• Kp is obtained equilibrium partial pressures expressed in atm are substituted into equilibrium –constant expression.

Kc NO2 2

N2O4 KpPNO2

2

PN2O4


15

• We can convert between Kc and Kp using an equation derived from PV = nRT:

For aA bB

Kp = Kc (RT) ∆n

∆n = moles gas products – moles of gas reactants∆n = b – aR = 0.08206 (L.atm)/K.mol)


16

• Heterogeneous Equilibrium:– When reacting species are in different phases– Solution and gas phases are included, solid and liquid

phases are excluded from the equilibrium equation because their concentrations “do not change.”

CaCO3(s) CaO(s) + CO2(g)

Kc = [CO2] because CaCO3 and CaO are solids.

Heterogeneous Equilibria

17

Thermal decomposition of calcium carbonate: CaCO3(s) <=> CaO(s) +CO2(g)

Heterogeneous Equilibria

18

Learning objectives: State generalizations concerning the composition of

equilibrium mixtures

Define reaction quotients. Relate reaction quotient to equilibrium constant.

Determine whether or not a system is at equilibrium for a given mixture of reactants and products. If it is not,determine the direction in which the reaction must go to achieve equilibrium.

Calculate the final concentrations of reactants and/orproducts from Kc and initial concentrations of reactantsand/or products.

Using Equilibrium Constants

19

Using Equilibrium Constants• The magnitude of equilibrium constant provides us with important

information about the composition of an equilibrium mixture.

• We can make the following generalizations concerning the composition of equilibrium mixtures:

If Kc > 103, products predominate over reactants. If Kc is very large, the reaction is said to proceed to completion.

If Kc is in the range 10–3 to 103, appreciable concentrations of both reactants and products are present.

If Kc < 10–3, reactants predominate over products. If Kc is very small, the reaction proceeds hardly at all.

20

Using the Equilibrium Constant

Kc = 4.2 x 10-48

2H2(g) + O2(g)2H2O(g)

(at 500 K)

Kc = 2.4 x 1047

2H2O(g)2H2(g) + O2(g)

(at 500 K)Kc = 57.0

2HI(g)H2(g) + I2(g)(at 500 K)

21

cC + dDaA + bB

Qc =[A]ta[B]tb[C]tc[D]tdReaction quotient:

The reaction quotient, Qc, is defined in the same way as the equilibrium constant, Kc, except that the concentrations in Qc are not necessarily equilibrium values.


22

• If Qc = Kc no net reaction occurs.

• If Qc < Kcnet reaction goes from left to right (reactants to products).

• If Qc > Kcnet reaction goes from right to left (products to reactants).

Using the Equilibrium Constant• To determine the direction in which the reaction will

proceed to achieve equilibrium, we compare the values of Qc and Kc:

23

• Predicting the direction of reaction. The direction of reaction depends on the relative values of Qc and Kc.

• If Qc < Kc, the reaction goes from left to right If Qc = Kc, the reaction is at equilibrium If Qc > Kc, the reaction goes from right to left

Using Equilibrium ConstantsUsing the reaction quotient

24

Steps to follow in calculating equilibrium concentrations from initial concentrations.

25


2HI(g)H2(g) + I2(g)

Kc =[H2][I2]

[HI]2

0.250 - 2xxxE

-2x+x+xC

I 0 0.2500

57.0 =x2

(0.250 - 2x)2

Substitute values into the equilibrium expression:

Set up a table:

26


Determine the equilibrium concentrations:

x = 0.0262x2

(0.250 - 2x)2

57.0 =

Solve for “x”:

HI: 0.250 - 2(0.0262) = 0.198 M

H2: 0.0262 M

I2: 0.0262 M

27

Learning objectives: State Le Chateliers’s Principle.

State the factors that alter the composition of an equilibrium mixture based on Le Chateliers’s Principle.

Explain the effect of the changes on reactant or product when a stress is applied to a system.

Determine the reaction direction when a system at equilibrium reacts to a stress applied to the system, including changes in concentrations, pressure and volume, or temperature.

Describe the effect of adding a catalyst to a system at equilibrium.

Le Châtelier’s Principle

28

Le Châtelier’s Principle: If a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress.

• The concentration of reactants or products can be changed.

• The pressure and volume can be changed.

• The temperature can be changed.


29

Changes in ConcentrationLe Châtelier’s Principle

Haber process for synthesis of ammonia.

N2(g) + 3 H2(g) 2 NH3(g) Kc = 0.291 at 700 K

Given an equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 at 700 K, what happens when the concentration of N2 is increased to 1.50 M?

Altering an Equilibrium Mixture: Concentration2NH3(g)N2(g) + 3H2(g)

31

– The concentration stress of an added reactant or product is relieved by reaction in the direction thatconsumes the added substance.

– The concentration stress of a removed reactant or product is relieved by reaction in the direction that replenishes the removed substance.

Le Châtelier’s PrincipleIn general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Châtelier’sprinciple predicts that

32

Changes in Pressure and VolumeIn general Le Châtelier’s principle predicts that

An increase in pressure by reducing the volume will bring about net reaction in the direction that decreasesthe number of moles of gas.

A decrease in pressure by enlarging the volume will bring about net reaction in the direction that increasesthe number of moles of gas.


33

N2(g) + 3 H2(g) 2 NH3(g) Kc = 0.291 at 700 K

Effect of pressure and volume

34

• the equilibrium constant for an exothermic reaction (negative H°) decreases as the temperatureincreases. Equilibrium shifts to form more reactants (reverse reaction).

• the equilibrium constant for an endothermic reaction (positive H°) increases as the temperature increases. Equilibrium shifts to form more product (forward reaction).

In general, when an equilibrium is disturbed by a change in temperature, Le Châtelier’s principle predicts that

Le Châtelier’s PrincipleChanges in Temperature

Altering an Equilibrium Mixture: Temperature2NH3(g)N2(g) + 3H2(g) H° = -2043 kJ

As the temperature increases, the equilibrium shifts from products to reactants.

36

Potential energy profiles for a reaction whose activation energy is lowered by the presence of a catalyst.

The activation energy for the catalyzed pathway (red curve) is lower than that for the uncatalyzed pathway (blue curve) by an amount ΔEa.

The catalyst lowers the activation energy barrier for the forward and reverse reactions by exactly the same amount.

Effect of a catalyst on equilibrium

The catalyst therefore accelerates the forward and reverse reactions by the same factor, and the composition of the equilibrium mixture is unchanged.

Chapter 5 Chemical Equilibrium

Documents

Transcript of Chapter 5 Chemical Equilibrium